$\dfrac{dy}{dx}=-\dfrac {xe^{- x^2}}y$ Which curve solves the differential equation and passes through the point $(0,1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y=-e^{- x^2}$ (Choice B) B $y=e^{- x^2/2}$ (Choice C) C $y=2-e^{- x^2/2}$ (Choice D) D $y=e^{- x^2}$ (Choice E) E $y=-e^{- x^2/2}$
Answer: The differential equation is separable. $\dfrac{dy}{dx}=-xe^{- x^2}\cdot\dfrac1y$ What does it look like after we separate the variables? $y\,dy=-xe^{- x^2}dx$ Let's integrate both sides of the equation. $\int y\,dy=\int -xe^{- x^2}dx$ We can use the substitution $u=-x^2$ with $du=-2x\,dx$ on the right side. What do we get? $\dfrac12 y^2=\dfrac12 e^{- x^2}+C$ What value of $C$ makes the solution curve pass through the point $(0,1)$ ? Let's substitute $x=0$ and $y=1$ into the equation and solve for $C$. $\begin{aligned} \dfrac12\cdot 1^2&=\dfrac12\cdot e^{-0^2}+C\\ \\ \\ \dfrac12&=\dfrac12+C\\ \\ \\ C&=0 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \dfrac12 y^2&=\dfrac12 e^{- x^2}+0\\ \\ \\ y^2&=e^{- x^2}\\ \\ y&=\pm\left( e^{- x^2} \right)^{1/2}\\ \\ y&=\pm e^{- x^2/2} \end{aligned}$ Which square root makes $y=1$ when $x=0$ ? The solution curve with the positive square root passes through the point $(0,1)$. $y=e^{- x^2/2}$